Euler's column theory – End conditions factor C According to Euler, the critical (crippling) load for a long column is p = π^2 E I / (C l^2). For a column with one end fixed and the other end free, what is the value of C?

Introduction / Context:Euler's buckling formula expresses the critical load in terms of column stiffness and effective length. Different end conditions change the effective length through the factor C.

Given Data / Assumptions:

• Long, slender column (Euler region) with modulus E and second moment I.
• Geometric length = l; one end fixed, other free (cantilever-like).
• Elastic stability analysis; initial imperfections neglected.

Concept / Approach:Effective length L_e captures boundary restraint. For fixed–free, L_e = 2 l. Substituting L_e in Euler's base form p = π^2 E I / L_e^2 gives p = π^2 E I / (4 l^2). Comparing with p = π^2 E I / (C l^2) yields C = 4.

Step-by-Step Solution:Identify end condition: fixed–free → L_e = 2 l.Base Euler: p = π^2 E I / L_e^2.Substitute: p = π^2 E I / (2 l)^2 = π^2 E I / (4 l^2).Match forms → C = 4.

Verification / Alternative check:Standard effective length factors: pinned–pinned C = 1; fixed–pinned C = 2; fixed–fixed C = 4; fixed–free C = 4 (because L_e = 2 l). Many texts tabulate these values consistently.

Why Other Options Are Wrong:(a) and (b) correspond to more restrained cases; (c) is for fixed–pinned; larger C reduces critical load, and fixed–free has the least capacity among common end conditions.

Common Pitfalls:Mixing up the two notations: some authors write p = π^2 E I / (K l)^2, where K is the effective length factor (K = 2 for fixed–free). Converting to the given form requires care.

Final Answer:4